Question
Question: A thin copper wire of length L increases in length by 1% when heated from temperature \({{\text{T}}_...
A thin copper wire of length L increases in length by 1% when heated from temperature T1toT2 what is the percentage change in area when a thin copper plate is heated having dimensions 2L !!×!! L is heated from T1toT2
a) 1%
b) 2%
c) 3%
d) 4%
Solution
- Hint: The material of the wire as well as the sheet is the same i.e. copper. Hence we can write the coefficient of linear expansion (α) is half the coefficient of superficial expansion( β), mathematically the relation can be written as, β=2α...(1). The percentage increase is basically the ratio of increase in dimension to the original dimension of the body. Hence we can relate the expansion of the two above bodies using the relation i.e. β=2α determine the percentage change in the area of the sheet.
Complete step-by-step solution
To begin with, let us write the equation of linear expansion and superficial expansion and express them in terms of the percentage increase in each of them.
The equation of linear expansion in a rod of length L is given by,
L=l(1+αΔT) where L is the length of the rod or the wire at the final change in temperature, l is the length of the rod or the wire at some initial temperature, α is the coefficient of linear expansion and ΔT is the change in temperature from the initial length of the rod to the final length of the rod. The above equation can also be written as,
L=l(1+αΔT)α=lΔTL−l...(2)
The term lL−l is analogous to the percentage increase in the length of the wire.
Now let us write the equation of superficial expansion.
A=a(1+βΔT) where A is the area of the sheet at final change in temperature, a is the area of the at some initial temperature, βis the coefficient of superficial expansion and ΔT is the change in temperature from the initial area of the sheet to the final area of the sheet. The above equation can also be written as,
A=a(1+βΔT)β=aΔTA−a....(3)
The term aA−a is analogous to the percentage increase in the area of the sheet.
The expansions of both sheet and the wire is taking place at the same change in temperature i.e. from T1toT2. Now let us write equation 1 and substitute equation 2 and 3 in that.
β=2αaΔTA−a=2(lΔTL−l)
!!Δ!! T is same for both the wire as well as the sheet, hence
aA−a=2(lL−l)aA−a=2(1%)aA−a=2%
Hence the percentage increase in the area of the sheet is 2%.
Note: We could use the equation 1 as the material of the sheet and the wire were the same i.e. copper. The term aA−a is taken as the percentage increase in the area for simplicity. The percentage increase is actually given by aA−a×100. We could just use the term aA−a as the increase in the area we equated this to the percentage increase in the length of the wire i.e. lL−l.